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                    <h1 class="description center-align post-title">「力扣」第 343 题：整数拆分</h1>
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                <h2 id="「力扣」第-343-题：“整数拆分”题解"><a href="#「力扣」第-343-题：“整数拆分”题解" class="headerlink" title="「力扣」第 343 题：“整数拆分”题解"></a>「力扣」第 343 题：“整数拆分”题解</h2><p>题解地址：<a href="https://leetcode-cn.com/problems/integer-break/solution/tan-xin-xuan-ze-xing-zhi-de-jian-dan-zheng-ming-py/" target="_blank" rel="noopener">“贪心选择”性质的简单证明、记忆化搜索、动态规划 （Python 代码、Java 代码）</a>。</p>
<p>说明：文本首发在力扣的题解版块，更新也会在第 1 时间在上面的网站中更新，这篇文章只是上面的文章的一个快照，您可以点击上面的链接看到其他网友对本文的评论。</p>
<p>传送门：<a href="https://leetcode-cn.com/problems/integer-break/" target="_blank" rel="noopener">343. 整数拆分</a>。</p>
<blockquote>
<p>给定一个正整数 n，将其拆分为至少两个正整数的和，并使这些整数的乘积最大化。 返回你可以获得的最大乘积。</p>
<p>示例 1:</p>
<p>输入: 2<br>输出: 1<br>解释: 2 = 1 + 1, 1 × 1 = 1。<br>示例 2:</p>
<p>输入: 10<br>输出: 36<br>解释: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36。<br>说明: 你可以假设 n 不小于 2 且不大于 58。</p>
</blockquote>
<h2 id="“贪心选择”性质的简单证明、记忆化搜索、动态规划-（Python-代码、Java-代码）"><a href="#“贪心选择”性质的简单证明、记忆化搜索、动态规划-（Python-代码、Java-代码）" class="headerlink" title="“贪心选择”性质的简单证明、记忆化搜索、动态规划 （Python 代码、Java 代码）"></a>“贪心选择”性质的简单证明、记忆化搜索、动态规划 （Python 代码、Java 代码）</h2><p>能够使用“贪心算法”完成的问题，严格上来说，是要证明这个问题具有“贪心选择”性质，但是如果一个问题，不能使用贪心算法，只需要举一个反例就可以了。</p>
<p>而证明一个问题具有“贪心选择”性质，通常是比较困难的。本篇就我对「力扣」第 343 题：“整数拆分”的贪心选择性质给予证明。</p>
<h3 id="方法一：贪心算法"><a href="#方法一：贪心算法" class="headerlink" title="方法一：贪心算法"></a>方法一：贪心算法</h3><p>（温馨提示：下面的幻灯片中，有几页上有较多的文字，可能需要您停留一下，可以点击右下角的后退 “|◀” 或者前进 “▶|” 按钮控制幻灯片的播放。）</p>
<p><img src="https://pic.leetcode-cn.com/8a4106141e27abd549e910ebc44ccc7e0af251e32ee509c7343db4d503718f19-343-1.png" alt="343-1.png">),<img src="https://pic.leetcode-cn.com/b520a336dc2ff0d60b9f801ea4aa2c76adfa17dc535a83c20fe1237d15c07c33-343-2.png" alt="343-2.png">),<img src="https://pic.leetcode-cn.com/b5c1887e62f85ac260dae3835cfca618745f427d343844d67e2655c0ebc6f10d-343-3.png" alt="343-3.png"></p>
<p><strong>分析题意</strong></p>
<p>题目中说，把正整数 $n$ 分解成若干个正整数加法因子，让我们求所有可能的加法因子相乘以后乘积的最大值。看示例 2 ，就很清楚题目的意思，这里就不多展开了。</p>
<p>本文对求解这个问题的贪心选择性质做了简单的归纳、证明。即我们要证明的是：</p>
<blockquote>
<p><strong>“贪心地”、“尽可能多”地分解出 $3$ 正整这个加法因子，就能够使得最终的乘积得到最大。</strong></p>
</blockquote>
<p><strong>从小规模数据找规律</strong></p>
<p>我们不妨从最小的正整数加法因子 $1$ 开始，到 $2$、$3$、$4$、…，依次分析，<strong>分解以后，乘积的大小，写几个出来，看看规律。</strong></p>
<ul>
<li>情况 1：如果分解式中含有 $1$ </li>
</ul>
<p>因为 $1 \times (n-1) = n-1 &lt; n$，明显，越分解乘积越小。</p>
<blockquote>
<p>结论1：$1$ 不能作为分解的正整数加法因子。</p>
</blockquote>
<ul>
<li>情况 2：如果分解式中含有 $2$ </li>
</ul>
<p>直觉上，分解以后相乘肯定比原来这个数要大。例如：$n = 19$ 时，$2 \times 17 = 34 &gt; 19$，越大的数越是如此，因此我们要关心的是边界。于是解不等式</p>
<p>$$2 \times  (n - 2) &gt; n$$</p>
<p>得 $n &gt; 4$。</p>
<blockquote>
<p>结论2：当 $n &gt; 4$ 的时候，分解出 $2$ 这个因子，与剩下的数 $(n - 2)$ 得到的乘积，就肯定超过 $n$ 了。</p>
</blockquote>
<ul>
<li>情况 3：如果分解式中含有 $3$ </li>
</ul>
<p>直觉上，也是分解以后相乘肯定比原来这个数要大。例如：$n = 19$ 时，$3 \times 16 = 48 &gt; 19$，越大的数越是如此，因此我们要关心的是边界。于是解不等式</p>
<p>$$3 \times  (n - 3) &gt; n$$</p>
<p>得 $n &gt; 4.5$。</p>
<blockquote>
<p>结论3：当 $n &gt; 4.5$ 的时候，分解出 $3$ 这个因子，与剩下的数 $(n - 3)$ 得到的乘积，就肯定超过 $n$ 了。</p>
</blockquote>
<ul>
<li>情况 4：如果分解式中含有 $4$ </li>
</ul>
<p>情况就有点意思了，我们就可以比较以下四者的乘积大小：</p>
<p>（1）$n$；</p>
<p>（2）$4 \times (n - 4)$；</p>
<p>（3）$2 \times 2 \times (n - 4)$；</p>
<p>（4）$1 \times 3 \times (n - 4)$，其实这都不用考虑，因为情况 1 的结论：$1$ 不能作为分解的正整数加法因子，因为它会使得乘积原来越小，下面我们马上还会分析到。</p>
<p>在 $n$ 比较大的时候，$4 \times (n - 4) &gt; n$ 是显然的，而 $4 \times (n - 4) = 2 \times 2 \times (n - 4)$，又由上面的第 1 点可知分解出 1 肯定使得乘积越来越小，而且很明显 $4 \times (n - 4) &gt; 1 \times 3 \times (n - 4)$。</p>
<p>结论：</p>
<blockquote>
<p>结论4：分解出 $4$，等价于分解出两个 $2$，因此，情况 2 就包含了情况 4 ，故没有必要考虑分解出正整数加法因子 $4$。</p>
</blockquote>
<ul>
<li>情况 5：如果分解式中含有 $5$ </li>
</ul>
<p>情况也有点意思了，我们就可以比较以下四者的乘积大小：</p>
<p>（1）$n$；</p>
<p>（2）$5 \times (n - 5)$；</p>
<p>（3）$2 \times 3 \times (n - 5)$；</p>
<p>（4）$1 \times 4 \times (n - 5)$；</p>
<p>在 $n$ 比较大的时候，$5 \times (n - 5) &gt; n$ 是显然的，而 $2 \times 3 \times (n - 5) &gt; 5  \times (n - 4)$，又由上面的第 1 点可知分解出 1 肯定使得乘积越来越小，而且很明显 $2 \times 3 \times (n - 5) &gt; 5  \times (n - 4) &gt; 1 \times 4 \times (n - 5)$。</p>
<blockquote>
<p>结论5：分解出 $5$ 与 $(n - 5)$ 的乘积还不如分解出 $2$ 、 $3$ 与 $(n-5)$ 的乘积，因此没有必要考虑分解出正整数加法因子 $5$。</p>
</blockquote>
<p>为了避免行文啰嗦，我就直接写最关键的部分了：</p>
<ul>
<li>情况 6：如果分解式中含有 $6$ </li>
</ul>
<p>$$6 \times (n-6)&lt; 2 \times 2 \times 2 \times (n - 6) &lt; 3 \times 3 \times (n - 6)$$</p>
<p>结论：</p>
<blockquote>
<p>分解出 $6$ 与 $(n - 6)$ 的乘积还不如分解出 $3$ 、 $3$ 与 $(n-6)$ 的乘积，因此没有必要考虑分解出正整数加法因子 $6$。</p>
</blockquote>
<p>写到这里，估计你也看出来了，继续分析下去的结论就是，</p>
<p>$$7 \times (n - 7) &lt; 2 \times 2 \times 3 \times (n - 7) $$</p>
<p>$$8 \times (n - 8) &lt; 2 \times 3 \times 3 \times (n - 8) $$</p>
<p>$$9 \times (n - 8) &lt; 3 \times 3 \times 3 \times (n - 9) $$</p>
<p>结论：</p>
<blockquote>
<p>结论6：$4$ 以上（包括 $4$）的正整数加法因子都不必考虑了，因为分解成它们的乘积，都小于分解出合适数量的 $2$ 和 $3$ 相乘以后的乘积。 </p>
</blockquote>
<p>根据以上分析，正整数加法因子只有 $2$ 和 $3$，在根据我们的直觉，感觉分解出 $3$ 这个正整数加法因子的乘积肯定比分解出 $2$ 这个正整数加法因子的乘积，还以 $n = 19$ 为例，$3 \times 16 = 48 &gt; 34 = 2 \times 17$，按照惯例，找边界，即解不等式</p>
<p>$$ 3 \times (n - 3) \ge 2 \times (n -2) $$</p>
<p>解得 $n \ge 5$。</p>
<blockquote>
<p>结论7：分解出 $3$ 比分解出 $2$ 好。</p>
</blockquote>
<p>综上所述：</p>
<blockquote>
<p><strong>我们应该尽可能分解出 $3$ ，直到最后剩下 $4$ 或者 $2$</strong>。</p>
</blockquote>
<p>下面我们分析最后剩下几就不能再分出 $3$ 了：</p>
<p>如果最后剩下 $7$ 的话，可以分解成两个 $3$、$3$、$2$；<br>如果最后剩下 $6$ 的话，可以分解成两个 $3$；<br>如果最后剩下 $5$ 的话，可以分解成 $3$ 和 $2$；<br>如果最后剩下 $4$ 的话，不能分解出 $3$ 了，因为最后剩下的正整数加法因子是 $1$，由结论 1 ，这是不允许的；<br>如果最后剩下 $3$ 的话，不能分解出 $2$ 了，因为剩下是 $1$，由结论 1 ，这是不允许的。</p>
<p>综上所述：</p>
<blockquote>
<p>在大于 $4$ 的前提下，尽可能分解出 $3$ 。当 $n \le 4$ 的时候，专门计算一下给出结论就可以了。</p>
</blockquote>
<p>以上分析其实并不难，关键要动手写一下，相信聪明的你一定会比我做得更好。</p>
<p><strong>参考代码 1</strong>：</p>
<p>Java 代码：</p>
<pre class="line-numbers language-Java"><code class="language-Java">public class Solution {
    public int integerBreak(int n) {
        if (n <= 2) {
            return 1;
        }
        if (n == 3) {
            return 2;
        }
        if (n == 4) {
            return 4;
        }
        // 接下来就是 n >= 5 的时候的逻辑了
        int res = 1;
        while (n > 4) {
            res *= 3;
            n -= 3;
        }
        res *= n;
        return res;
    }
}<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>
<p>Python 代码：</p>
<pre class="line-numbers language-Python"><code class="language-Python">class Solution:
    def integerBreak(self, n):
        if n == 2:
            return 1
        if n == 3:
            return 2
        if n == 4:
            return 4
        res = 1
        while n > 4:
            res *= 3
            n -= 3

        res *= n
        return res<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>
<h3 id="方法二：“记忆化搜索”与“动态规划”"><a href="#方法二：“记忆化搜索”与“动态规划”" class="headerlink" title="方法二：“记忆化搜索”与“动态规划”"></a>方法二：“记忆化搜索”与“动态规划”</h3><p>思路：先研究递归结构，发现有大量重叠子问题，再实现“记忆化搜索”，最后实现使用“动态规划”。即先“自顶向下”思考，再“自底向上”实现。</p>
<p><img src="https://pic.leetcode-cn.com/6adc05350bfd3700fba01636a5a72b1e2474d643b379acf264b60c9426b734fe-image.png" alt="image.png"></p>
<p>注意：<strong>对于每一个状态而言，还要再比较“不再继续分割”和“继续分割”，取当中的最大值</strong>，将 $n$ 进行分解的时候，以 $8$ 为例：$1$ 与 $7$ 是一个解，$1$ 与 $7$ 的分解的结果也是一个解。</p>
<p><strong>参考代码 2</strong>：记忆化搜索</p>
<p>Python 代码：</p>
<pre class="line-numbers language-Python"><code class="language-Python">class Solution:
    def __init__(self):
        self.memo = []

    def integerBreak(self, n):
        self.memo = [-1 for _ in range(n + 1)]
        self.memo[0] = 1
        self.memo[1] = 1
        return self.__dfs(n)

    def __dfs(self, n):
        if n == 1:
            return 1
        if self.memo[n] == -1:
            res = 0
            for i in range(1, n):
                res = max(res, i * (n - i), i * self.__dfs(n - i))
            self.memo[n] = res
        return self.memo[n]<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>
<p>Java 代码：</p>
<pre class="line-numbers language-Java"><code class="language-Java">public class Solution {

    private int[] memory;

    public int integerBreak(int n) {
        assert n >= 2;
        memory = new int[n + 1];
        memory[0] = 0;
        memory[1] = 1;
        for (int i = 2; i < n + 1; i++) {
            memory[i] = -1;
        }
        int res = breakInteger(n);
        return res;
    }


    // 将 n 进行分割得到的乘积最大值
    private int breakInteger(int num) {
        if (num == 1) {
            return 1;
        }
        if (memory[num] == -1) {
            int res = 0; // 这个初始值可以设置为 0 吗，1 行不行？
            for (int i = 1; i < num; i++) {
                // 关键之处：状态转移方程，其中 i * (num - i) 这一步很关键，千万不能漏掉
                res = max3(res, i * (num - i), i * breakInteger(num - i));
            }
            memory[num] = res;
        }
        return memory[num];
    }

    private int max3(int num1, int num2, int num3) {
        int temp = Integer.max(num1, num2);
        return Integer.max(temp, num3);
    }

    public static void main(String[] args) {
        Solution2 solution = new Solution2();
        int max = solution.integerBreak(9);
        System.out.println(max);
    }
}<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>
<p><strong>参考代码 3</strong>：动态规划</p>
<p>Java 代码：</p>
<pre class="line-numbers language-Java"><code class="language-Java">/**
 * 动态规划的解法
 * Created by liwei on 17/10/3.
 */
public class Solution3 {

    private int[] memory;

    public int integerBreak(int n) {
        memory = new int[n + 1];
        memory[0] = 0;
        memory[1] = 1;
        for (int i = 2; i <= n; i++) {
            int maxValue = -1;
            for (int j = 1; j <= i - 1; j++) {
                maxValue = max3(maxValue, j * (i - j), j * memory[i - j]);
            }
            memory[i] = maxValue;
        }
        return memory[n];
    }

    private int max3(int num1, int num2, int num3) {
        int temp = Integer.max(num1, num2);
        return Integer.max(temp, num3);
    }

    public static void main(String[] args) {
        Solution3 solution = new Solution3();
        int max = solution.integerBreak(9);
        System.out.println(max);
    }
}<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>
<p>Python 代码：</p>
<pre class="line-numbers language-Python"><code class="language-Python">class Solution:
    def integerBreak(self, n):
        dp = [1 for _ in range(n + 1)]
        for i in range(2, n + 1):
            for j in range(1, i):
                dp[i] = max(dp[i], j * dp[i - j], j * (i - j))
        return dp[n]<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>

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